hernandez (ejh742) – homework 13 – Turner – (58120)
1
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printout
should
have
12
questions.
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before answering.
001
10.0 points
A car with mass 665 kg passes over a bump
in a road that follows the arc of a circle of
radius 37
.
4 m as shown in the figure.
The acceleration of gravity is 9
.
8 m
/
s
2
.
37
.
4 m
v
665 kg
What is the maximum speed the car can
have as it passes the highest point of the
bump before losing contact with the road?
Correct answer: 19
.
1 m
/
s.
Explanation:
At the highest point, we have
m g
 N
=
m v
2
r
,
where
N
is the normal force.
To get the
maximum speed, we need
N
= 0
.
Therefore,
v
max
=
√
g r
=
radicalBig
(9
.
8 m
/
s
2
) (37
.
4 m)
=
19
.
1 m
/
s
.
002
(part 1 of 2) 10.0 points
A small sphere of mass
m
is connected to
the end of a cord of length
r
and rotates
in a
vertical
circle about a fixed point O. The
tension force exerted by the cord on the sphere
is denoted by
T
.
r
O
θ
What is the correct equation for the forces
in the radial direction when the cord makes
an angle
θ
with the vertical?
1.
T

m g
sin
θ
=

m v
2
r
2.
T

m g
sin
θ
= +
m v
2
r
cos
θ
3.
None of these
4.
T

m g
cos
θ
= +
m v
2
r
correct
5.
T
+
m g
cos
θ
= +
m v
2
r
6.
T

m g
sin
θ
= +
m v
2
r
7.
T

m g
sin
θ
=

m v
2
r
tan
θ
8.
T

m g
sin
θ
= +
m v
2
r
tan
θ
9.
T
+
m g
sin
θ
= +
m v
2
r
Explanation:
O
θ
θ
mg
T
The centripetal force is
F
c
=
m v
2
r
.
This centripetal force is provided by the ten
sion force and the radial component of the
weight.
In this case, they are in opposite
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hernandez (ejh742) – homework 13 – Turner – (58120)
2
direction, so
F
c
=
m v
2
r
=
T

m g
cos
θ .
003
(part 2 of 2) 10.0 points
What is the magnitude of the total accel
eration? You may want to first find both the
radial and the tangential component of the
acceleration.
1.

vectora

=
radicalBigg
parenleftbigg
T
m
parenrightbigg
2

2
g
cos
θ
T
m
+
g
2
cor
rect
2.

vectora

=
radicalBigg
parenleftbigg
T
m
parenrightbigg
2
+
g
2
3.

vectora

=
radicalBigg
parenleftbigg
T
m
parenrightbigg
2
+ 2
g
cos
θ
T
m
+
g
2
4.

vectora

=
radicalBigg
parenleftbigg
T
m
parenrightbigg
2
sin
2
θ
+
g
2
cos
2
θ
5.

vectora

=
radicalBigg
parenleftbigg
T
m
parenrightbigg
2
cos
2
θ
+
g
2
sin
2
θ
6.

vectora

=
radicalBigg
parenleftbigg
T
m
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 Spring '08
 Turner
 Physics, Acceleration, Force, Mass, Work, Correct Answer

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